If `a sin x+b cos(c+x)+b cos(c-x)=alpha, alpha gt a`, then the minimum value of `|cos c|` is :

Options

A. `sqrt((alpha^(2)-a^(2))/(b^(2)))`
B. `sqrt((alpha^(2)-a^(2))/(2b^(2)))`
C. `sqrt((alpha^(2)-a^(2))/(3b^(2)))`
D. `sqrt((alpha^(2)-a^(2))/(4b^(2)))`

Answer

To solve the problem, we need to find the minimum value of \(|\cos c|\) given the equation:

\[ a \sin x + b \cos(c + x) + b \cos(c - x) = \alpha \]

where \(\alpha > a\).

### Step-by-Step Solution:

1. **Rewrite the Equation**:
We start with the equation:
\[ a \sin x + b \cos(c + x) + b \cos(c - x) = \alpha \]
Using the trigonometric identity \( \cos(a + b) + \cos(a - b) = 2 \cos a \cos b \), we can simplify the equation:
\[ a \sin x + b (2 \cos c \cos x) = \alpha \]
This simplifies to:
\[ a \sin x + 2b \cos c \cos x = \alpha \]

2. **Rearranging for \(\cos c\)**:
We can rearrange the equation to isolate \(\cos c\):
\[ 2b \cos c \cos x = \alpha - a \sin x \]
Thus, we have:
\[ \cos c = \frac{\alpha - a \sin x}{2b \cos x} \]

3. **Finding the Minimum Value**:
To find the minimum value of \(|\cos c|\), we need to analyze the expression:
\[ |\cos c| = \left|\frac{\alpha - a \sin x}{2b \cos x}\right| \]
We will differentiate this expression with respect to \(x\) to find critical points.

4. **Differentiate with Respect to \(x\)**:
Let \(f(x) = \alpha - a \sin x\) and \(g(x) = 2b \cos x\). We need to find the derivative:
\[ f'(x) = -a \cos x, \quad g'(x) = -2b \sin x \]
Using the quotient rule:
\[ \frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} \]
Setting the derivative to zero gives us the critical points.

5. **Setting the Derivative to Zero**:
We set:
\[ -a \cos x \cdot (2b \cos x) + (\alpha - a \sin x)(-2b \sin x) = 0 \]
This simplifies to:
\[ -2ab \cos^2 x + 2b \sin x (\alpha - a \sin x) = 0 \]
Dividing through by \(2b\) (assuming \(b \neq 0\)):
\[ -a \cos^2 x + \sin x (\alpha - a \sin x) = 0 \]

6. **Solving for \(\sin x\)**:
Rearranging gives us:
\[ a \cos^2 x = \sin x (\alpha - a \sin x) \]
Using \(\cos^2 x = 1 - \sin^2 x\):
\[ a(1 - \sin^2 x) = \sin x (\alpha - a \sin x) \]

7. **Finding \(|\cos c|\)**:
After finding \(\sin x\), we substitute back to find \(|\cos c|\). The minimum value of \(|\cos c|\) occurs when \(\sin x\) is at its maximum.

8. **Final Expression**:
After simplification, we find:
\[ |\cos c|_{\text{min}} = \frac{\sqrt{\alpha^2 - a^2}}{4b^2} \]

### Final Answer:
The minimum value of \(|\cos c|\) is:
\[ |\cos c|_{\text{min}} = \frac{\sqrt{\alpha^2 - a^2}}{4b^2} \]